Gödel's theorems, Löb's theorem, difference between "proves" and "implies"

Question

Layman reading up on Gödel's theorems. I think I have some basic idea of how it all works in the abstract, but I'm still having a hard time distinguishing between (or even counting) the concepts involved. I initially thought I was quite comfortable with the distinctions between

• "$X$ is true" versus "$X$ is provable within $\mathsf{S}$" (i.e. "$X$ is a theorem of $\mathsf{S}$");
• $X\rightarrow Y$ "$X$ implies $Y$" versus $[\mathsf{S} + X]\vdash Y$ "$X$ proves $Y$ within $\mathsf{S}$" (i.e. "$X\supset Y$ is a theorem of $\mathsf{S}$").

But after writing up this question, I'm no longer even sure of that. I'm still having a hard time with two things, which are probably one thing:

(1) Löb's theorem, IIUC, says that $$\mathsf{PA}\vdash[[\mathsf{PA}\vdash X]\supset X]\rightarrow[\mathsf{PA}\vdash X].$$ But surely $[\mathsf{PA}\vdash X]\supset X]$ is always the case within $\mathsf{PA}$, isn't it? That is, whenever $\mathsf{PA}\vdash X$ — whenever we have a chain of formulas leading (by $\mathsf{PA}$'s rules of deduction) from one-or-more-axioms-of-$\mathsf{PA}$ to $X$ — then that is exactly what it means for $X$ to be a theorem of $\mathsf{PA}$, isn't it?

My vague solution here is that there must be a significant difference between the two theorems $\mathsf{PA}\vdash X$ "$\mathsf{PA}$ proves $X$" and $\mathsf{PA}\supset X$ "$\mathsf{PA}$ implies $X$." From the latter, we can obviously conclude $X$ is a theorem of $\mathsf{PA}$: $$ \begin{aligned} &\mathsf{PA}\vdash\mathsf{PA}\quad&&\text{(axiomatically)}\\ &\mathsf{PA}\vdash[\mathsf{PA}\supset X]\quad&&\text{(premise)}\\ &\mathsf{PA}\vdash X\quad&&\text{(by lines 1 and 2, modus ponens)} \end{aligned} $$ but maybe from the former we cannot?: $$ \begin{aligned} &\mathsf{PA}\vdash\mathsf{PA}\quad&&\text{(axiomatically)}\\ &\mathsf{PA}\vdash[\mathsf{PA}\vdash X]\quad&&\text{(premise)}\\ &\mathsf{PA}\vdash^? X\quad&&\text{(“by lines 1 and 2,” but no rule of inference within $\mathsf{PA}$ actually gets us here)} \end{aligned} $$

(2) Gödel says that the system $\mathsf{T}=[\mathsf{PA}+\neg Con(\mathsf{PA})]$ is consistent. But surely the axiom $\neg Con(\mathsf{PA})$ means no more or less than $\forall X:[\mathsf{PA}\vdash X]$. Then we have $$ \begin{aligned} &\mathsf{T}\vdash[\forall X:[\mathsf{PA}\vdash X]]\quad&&\text{(axiomatically)}\\ &\mathsf{T}\vdash[\mathsf{PA}\vdash [0\not=0]]\quad&&\text{by line 1, substitution of $[0\not=0]$ for $X$}\\ &\mathsf{T}\vdash\mathsf{PA}\quad&&\text{axiomatically}\\ &\mathsf{T}\vdash^?[0\not=0]\quad&&\text{“by lines 2 and 3,” but how?}\\ \end{aligned} $$ I'm having a hard time seeing how we can't take the provability of the premise (line 3), combine it with the fact that a deduction from premise to conclusion exists (line 2), and deduce the conclusion (line 3). I suppose the problem is again that $\vdash$ "proves" is a different symbol from $\supset$ "implies," and modus ponens is allowed to work only on the latter, not on the former...? ...But this isn't satisfying, because it seems like a silly restriction on $\mathsf{PA}$'s modus ponens in particular. If a formalization of math is allowed to take obvious truths as axioms, I don't see why we wouldn't choose to take that obvious truth as an axiom! (It is obviously true of math that whatever is provable, is true.) Consider a system that's just like $\mathsf{PA}$ but has one additional rule of inference: $$\mathsf{U} = [\mathsf{PA} + \forall X:[\mathsf{U}\vdash X]\supset X];$$ then... well, actually, $\forall X:[\mathsf{U}\vdash X]\supset X$ is just another way of writing $Con(\mathsf{U})$, isn't it? So $\mathsf{U}$ would be able to prove its own consistency (trivially, from the axiom); and thus (by Gödel) $\mathsf{U}$ would have to be inconsistent?

Basically, if I'm correct at all that it's important to distinguish $\vdash$ from $\supset$, then I think I'm struggling to intuitively grasp the idea of how a system $\mathsf{T}$ could ever come up with $X\vdash Y$ as a theorem, other than by also having $X\supset Y$ as a theorem. $X\supset Y$ would be a "constructive proof" that $X$ proves $Y$; $X\vdash Y$ without $X\supset Y$ would be a "nonconstructive proof" that $X$ proves $Y$. But I'm struggling to imagine how one would prove $X\vdash Y$ as a theorem of $\mathsf{T}$ (other than literally starting with it as an axiom). If this is my problem, maybe someone could give a toy example? Or, if it sounds like I've got a different problem, please point it out!

SAME-DAY EDIT: Actually, it occurs to me that in this context it may be really significant exactly which formal sentence we use to represent the axiom colloquially expressed as $\neg Con(\mathsf{T})$. We could translate that into the axiom $\mathsf{T}\vdash0\not=0$, or the axiom $\neg\forall X:[\mathsf{T}\vdash X]\supset X$, or (maybe?) even the axiom "a specific Gödel sentence for $\mathsf{T}$." Or is it already that case that any formula of $\mathsf{T}$ that seems to involve the $\vdash$ symbol is really just a colloquialism for "some complicated statement about the Gödel numbers of proofs w.r.t. some specific Gödelization of $\mathsf{T}$" and $\mathsf{T}$ per se can't talk about provability at all? IOW, I may already have accidentally slipped one level of abstraction.

Answer 1

But surely $[[\mathsf{PA}\vdash X]\supset X]$ is always the case within , isn't it?

Not internally, no! For example, "as far as $\mathsf{PA}$ knows" (= consistently with $\mathsf{PA}$, by the second incompleteness theorem) it could be the case that $\mathsf{PA}\vdash\perp$. But $\mathsf{PA}$ outright disproves $\perp$. In particular, we (hopefully!) have $$\mathsf{PA}\not\vdash [\mathsf{PA}\vdash \perp]\supset \perp$$ since otherwise we'd get an inconsistency in $\mathsf{PA}$ by the second incompleteness theorem.

The issue is that you're thinking of the provability relation "too correctly." It helps to think semantically. Suppose $\mathcal{M}$ were a model of $\mathsf{PA}+\neg\mathsf{Con(PA)}$ (this theory is consistent by the second incompleteness theorem, and so it has a model by the completeness theorem). Such an $\mathcal{M}$ is quite complicated, but at heart it's really just a discretely ordered semiring with certain technical features. There is a particular relation $R(x,y)$, defined in terms of the semiring structure, which in the standard model corresponds to "$x$ is a code for a $\mathsf{PA}$-proof of the sentence coded by $y$." Whenever we talk about provability within $\mathsf{PA}$ itself, what we're really doing is using this relation. There is no reason to expect $R$ to continue to match up nicely with "true provability" once our model is nonstandard. In particular, $\mathcal{M}$ contains an element which it "thinks" is a code for a $\mathsf{PA}$-proof of $\perp$, but such an element is not a true natural number (= doesn't correspond to an actual proof); phrased in less philosophically loaded terms, such an element is in the non-Archimedean part of $\mathcal{M}$.

It may help at this point to replace the theory $\mathsf{PA}$ (about which we have many intuitions which may confound the picture) with the theory $T:=\mathsf{PA+\neg Con(PA)}$. Crucially we have $$T\vdash [T\vdash \perp]\wedge\neg\perp,$$ so the $T$-analogue of the scheme you're asking about is not only not-$T$-provable (which parallels the $\mathsf{PA}$-setting) but is actually $T$-disprovable (which is not the case for $\mathsf{PA}$).